Product Failure

I’ve been taking a grad course in statistics this semester and so I’ve been thinking about all sorts of real world examples of math, including the classic product-failure example that’s a mainstay of most stat classes.

One of the simplest continuous distributions is the exponential distribution which is a pretty decent way to model product failures. The probability of failure f(t) after time t is given by

f(t)=\frac{1}{\lambda}e^{-t/\lambda}.

I read this great article about product failure and testing in Wired this week. I encourage you to check it out. Read the last page of the article especially, where it talks about how cutting-edge companies are modeling minute variations in materials using an electron microscope and some statistics. Instead of actually testing the product over and over again using a fatigue machine, they can create surprisingly accurate models of the materials using computers. Prior to this, the behavior of materials was somewhat unpredictable.

Of course I was excited to see this figure in the article, which shows the Weibull distribution modeling failures of steel bars in a fatigue machine.

The Weibull distribution, unlike the exponential distribution, takes the age of a product into account. If the parameter k is greater than zero, than the rate of product failure increases with time. The probability of failure f(t) after time t is

f(t)=\frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1}e^{-(t/\lambda)^k}.

The first obvious thing to note is that the exponential distribution is just a special case of the Weibull distribution, with k=1. The next thing to say is that this distribution is single-peaked. So how is the above a Weibull distribution? The article says it is, but I think it might be a linear combination of two Weibull distributions, don’t you? Whatever–normalize, and you’ve got yourself a probability distribution.

[pun warning!]

The real question is, if this is TWO Weibulls, would you settle for the lesser of two Weibulls?

Sorry. I had to.

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